The perimeter of an equilateral triangle is 32 centimeters. How do you find the length of an altitude of the triangle?

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The perimeter of an equilateral triangle is 32 centimeters. How do you find the length of an altitude of the triangle?

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Answer 1 · 2016-02-16T22:45:19

Calculated "from grass roots up"

$h = 5 \frac{1}{3} \times \sqrt{3}$ $h= 5 1/3 xx sqrt(3)$ as an 'exact value'

Explanation:

Tony B

$By using fractions when able you do not introduce error$ $color(brown)("By using fractions when able you do not introduce error")$ $and some times things just cancel out or simplify!!!$ $color(brown)("and some times things just cancel out or simplify!!!"$

Using Pythagoras

$h^{2} + {(\frac{a}{2})}^{2} = a^{2}$ $h^2 + (a/2)^2 =a^2$ ...........................(1)

So we need to find $a$ $a$

We are given that the perimeter is 32 cm

So $a + a + a = 3 a = 32$ $a+a+a = 3a = 32$

So $a = \frac{32}{3} s o a^{2} = {(\frac{32}{3})}^{2}$ $" "a=32/3" " so " " a^2=(32/3)^2$

$\frac{a}{2} = \frac{1}{2} \times \frac{32}{3} = \frac{32}{6}$ $a/2" " = " "1/2xx32/3" "=" "32/6$

${(\frac{a}{2})}^{2} = {(\frac{32}{6})}^{2}$ $(a/2)^2= (32/6)^2$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Substituting these value into equation (1) gives

$h^{2} + {(\frac{a}{2})}^{2} = a^{2} \to h^{2} + {(\frac{32}{6})}^{2} = {(\frac{32}{3})}^{2}$ $h^2+(a/2)^2 =a^2" "-> " "h^2+(32/6)^2=(32/3)^2$

$h = \sqrt{{(\frac{32}{3})}^{2} - {(\frac{32}{6})}^{2}}$ $h= sqrt( (32/3)^2-(32/6)^2)$

There is a very well known algebra method hear where if we have
$(a^{2} - b^{2}) = (a - b) (a + b)$ $(a^2-b^2) = (a-b)(a+b)$

also $\frac{32}{3} = \frac{64}{6}$ $32/3= 64/6$ so we have

$h = \sqrt{(\frac{64}{6} - \frac{32}{6}) (\frac{64}{6} + \frac{32}{6})}$ $h= sqrt( (64/6-32/6)(64/6+32/6)$

$h = \sqrt{(\frac{32}{6}) (\frac{96}{6})}$ $h= sqrt( (32/6)(96/6)$

$h = \sqrt{\frac{1}{6^{2}} \times 32 \times 96}$ $h= sqrt( 1/6^2xx32xx96$

By looking at the 'factor tree' we have
$32 \to 2 \times 4^{2}$ $32 -> 2xx4^2$
$96 \to 2^{2} \times 2^{2} \times 3 \times 2$ $96-> 2^2xx2^2xx3xx2$

giving:

$h = \sqrt{\frac{1}{6^{2}} \times 2^{2} \times 2^{2} \times 2^{2} \times 4^{2} \times 3}$ $h= sqrt( 1/6^2xx2^2xx2^2 xx2^2xx4^2xx3)$

$h = \frac{1}{6} \times 2 \times 2 \times 2 \times 4 \times \sqrt{3}$ $h= 1/6xx2xx2xx2xx4xxsqrt(3)$

$h = \frac{32}{6} \sqrt{3}$ $h=32/6 sqrt(3)$

$h = 5 \frac{1}{3} \times \sqrt{3}$ $h= 5 1/3 xx sqrt(3)$ as an 'exact value'

Answer 2 · 2016-02-16T23:16:28

Calculated using a quicker method: By ratio

$h = 5 \frac{1}{3} \sqrt{3}$ $h= 5 1/3 sqrt(3)$
$How is that for shorter!!!!$ $color(red)("How is that for shorter!!!!")$

Explanation:

Tony B

If you had an equilateral triangle of side length 2 then you would have the condition in the above diagram.
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
We know that the perimeter in the question is 32 cm. So each side is of length:

$\frac{32}{3} = 10 \frac{2}{3}$ $32/3 =10 2/3$

So $\frac{1}{2}$ $1/2$ of one side is $5 \frac{1}{3}$ $5 1/3$

So by ratio, using the values in this diagram to those in my other solution we have:

$\frac{10 \frac{2}{3}}{2} = \frac{h}{\sqrt{3}}$ $(10 2/3)/2 = h/(sqrt(3))$

so $h = (\frac{1}{2} \times 10 \frac{2}{3}) \times \sqrt{3}$ $h= (1/2 xx 10 2/3) xx sqrt(3)$

$h = 5 \frac{1}{3} \sqrt{3}$ $h= 5 1/3 sqrt(3)$

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The perimeter of an equilateral triangle is 32 centimeters. How do you find the length of an altitude of the triangle?

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